A thin disc of radius b = 2 a has a concentric hole of radius a in it (see figure). It carries uniform surface charge σ on it. If the electric field on its axis at a height h h < < a from its centre is given as C h then the value of C is

Question

A thin disc of radius b = 2 a has a concentric hole of radius a in it (see figure). It carries uniform surface charge &#963; on it. If the electric field on its axis at a height h h &#60; &#60; a from its centre is given as C h then the value of C is

Aicharya · Accepted Answer

&#8757; At the axial point of a uniformly charged disc electric field is given by, E = &#963; 2 &#1013; 0 1 - c o s &#952; By superposition principle, when inner disc is removed , then, electric field due to remaining disc is, E = &#963; 2 &#1013; 0 &#160; 1 - c o s &#952; 2 - 1 - c o s &#952; 1 = &#963; 2 &#1013; 0 c o s &#952; 1 - c o s &#952; 2 = &#963; 2 &#1013; 0 h h 2 + a 2 &#160; - h h 2 + b 2 = &#963; 2 &#1013; 0 h a 1 + h 2 a 2 - h b 1 + h 2 b 2 &#160; &#8757; h &#160; &#8810; a and b. &#8756; E = &#963; 2 &#1013; 0 h a - h b = &#963; 2 &#1013; 0 h a - h 2 a = &#963; h 4 &#1013; 0 a &#8658; C = &#963; 4 a &#1013; 0 .

A thin disc of radius b = 2 a has a concentric hole of radius a in it (see figure). It carries uniform…

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